# B Are differential angles vectors?

1. Mar 11, 2016

### Cozma Alex

Because on the book it is said that little angles are vectors but my question is:
Are they vectors at all or they are scalar and we assign them a direction by multiplying them by a versor? The same for angular velocity, is it a vector at all or we made it a vector for making the right hand rule work?

2. Mar 11, 2016

### alw34

A scalar has magnitude only, such as temperatures at different points in a room.

A vector has not only magnitude but direction, say the flow of water during the draining of a bath tub.

So if I want to give you directions from you to me, saying I am five miles away [magnitude] is not enough; I need to tell you 'fives miles away to the west' , for example, using a compass as the scale.

https://en.wikipedia.org/wiki/Euclidean_vector

3. Mar 11, 2016

### drvrm

finite angular displacements are not vectors as they do not obey the the rule that A + B = B+ A ,
but in infinitesimal rotations the displacements can have a direction along the axis of rotations - clockwise/anticlockwise rotation can have two directions
and the time rate of change of angles do provide a vector called angular velocity- no doubt the rotations in general can have three components like the normal vectors or generalized rotations......

4. Mar 11, 2016

### alw34

unlikely. best to give actual quotes so we know what your reference is.

'Little angles' are angles, not vectors, right? For one thing an angle is usually not referenced to a coordinate frame of reference, like a graph plot. Vectors have direction because they have a specific direction with respect to a frame.

5. Mar 13, 2016

### vanhees71

I don't know, what you mean by "little vectors", but an infinitesimal rotation can be written in terms of an axial vector $\delta \vec{\varphi}$:
$$\delta \vec{V}=\delta \vec{\varphi} \times \vec{V}.$$
To see this look at a rotation around an axis $\vec{n}$ (right-hand rule!) with an angle $\phi$. If you take the $z$ axis of a Cartesian righthanded coordinate system then it's described by the matrix
$$\hat{D}=\hat{D}_{\vec{n}}(\varphi)=\begin{pmatrix} \cos \varphi & -\sin \varphi &0 \\ \sin \varphi & \cos \varphi & 0 \\ 0 & 0 & 1 \end{pmatrix},$$
i.e.,
$$\hat{D} \vec{V}=V_z \vec{e}_z + \cos \varphi (V_x \vec{e}_x + V_y \vec{e}_y)+\sin \varphi (-V_y \vec{e}_x+V_x \vec{e}_y).$$
On the other hand we have
$$\vec{e}_z \times \vec{V}=\vec{n} \times \vec{V}=-V_y \vec{e}_x+ V_x \vec{e}_y, \quad \vec{n} \times (\vec{n} \times \vec{V}) = -V_x \vec{e}_x-V_y \vec{e}_y,$$
from which we find
$$\hat{D} \vec{V}=\vec{n} (\vec{n} \cdot \vec{V})-\cos \varphi \vec{n} \times (\vec{n} \times \vec{V}) + \sin \varphi \vec{n} \times \vec{V}.$$
For a small angle $\delta \varphi$ this implies
$$\hat{D} \vec{V}=\vec{n} (\vec{n} \cdot \vec{V}) - \vec{n} \times (\vec{n} \times \vec{V}) +\delta \varphi \vec{n} \times \vec{V} + \mathcal{O}(\delta \varphi^2).$$
Now we have
$$\vec{n} \times (\vec{n} \times \vec{V}) =\vec{n} (\vec{n} \cdot \vec{V})-\vec{V} (\vec{n} \cdot \vec{n}) = \vec{n} (\vec{n} \cdot \vec{V})-\vec{V}$$
and thus finally
$$\hat{D} \vec{V}=\vec{V} +\delta \varphi \vec{n} \times \vec{V} + \mathcal{O}(\delta \varphi^2).$$